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The first irregular prime is 37. Does FLT(37)

$$x^{37} + y^{37} = z^{37}$$

have any solutions in the ring of integers of $\mathbb Q(\zeta_{37})$, where $\zeta_{37}$ is a primitive 37th root of unity?

Maybe it's not true, but how could I go about finding a counter-example? (for any cyclotomic ring, not necessarily 37)

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quanta
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    The question of for which $n \in \mathbb{Z}^+$ the analogue of FLT holds in the cyclotomic ring $\mathbb{Z}[\zeta_n]$ is surely unsolved in general: this is really hard! Just as surely there must be some work on special cases (possibly including $n = 37$; my blind guess is that's close to the boundary of what can be done with ad hoc techniques). You might want to ask this question on Math Overflow instead, the better to get a sense of exactly how much the experts know. – Pete L. Clark Feb 13 '11 at 21:23
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    Prior to Wiles' proof of FLT, one statement that I often heard was that it was known that for each $n > 2$, there can only be finitely many coprime solutions to FLT(n). This is actually due to Falting's theorem, which says that there are only finitely many rational points on an algebraic curve of genus greater than 1. This statement holds in any number field. So, we do know that there can only be finitey many solutions (up to rescaling) to FLT(37) in any number field and, in particular, in $\mathbb{Q}(\zeta_{37})$. Unfortunately, Falting's theorem is not effective, meaning that it does not... – George Lowther Feb 13 '11 at 21:32
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    ...provide you with a bound on the largest possible solution so, as far as I know, it doesn't reduce the problem to a finite search. – George Lowther Feb 13 '11 at 21:33
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    Also, Springer Online Reference (http://eom.springer.de/i/i052670.htm) mentions Vandiver's test as a way of determining that FLT(37) holds. It does not mention if that means it holds in the cyclotomic numbers though, and searching for "Vandiver's test" is practically a googlewack (that link is the only result, if I search for "Vandiver's test" in quotes). – George Lowther Feb 13 '11 at 21:37
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    There's no solutions to the first case of FLT(37) in $\mathbb{Q}(\zeta_{37})$, because $2^{37}=76\not=2$ mod $37^2$. See here: http://dx.doi.org/10.1070/IM1999v063n05ABEH000262 – George Lowther Feb 13 '11 at 21:55
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    @George: sure, Faltings' theorem applies to the Fermat curve for any $n > 3$ (note: but not for $n = 3$ -- this is an elliptic curve) over any number field $K$. So there are finitely many rational points on all these Fermat curves...but this is pretty far away from knowing whether there are more than $3$ points! – Pete L. Clark Feb 13 '11 at 22:21
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    I have posted an updated version of this question to mathoverflow. – quanta Feb 20 '11 at 21:13
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    You should ask the more general question of solving Diophantine equations on arbitrary number fields, right? – awllower Apr 06 '11 at 04:48
  • @awllower, I have found some counterexamples to FLT in various numbers but the reason for asking specifically about FLT(p) in cyclotomic field p is that I am very often seeing FLT(p) proved impossible in integers by proving it impossible in cyclotomic p integers. Basically there seems to be something happening behind the scenes.. – quanta Apr 06 '11 at 10:22
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    @quanta: Is it helpful to consider abelian fields instead? – awllower Apr 07 '11 at 13:05
  • @awllower, at present I have no ideas how to approach this problem. – quanta Apr 07 '11 at 13:21

1 Answers1

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This question was answered in mathoverflow. I am writing this to close up this question and making this answer as a community wiki according to MSE's guidelines. The answer is due to Tauno Metsänkylä.

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